Calculus and vectors #rvc
Time-dependent vectors can be differentiated in exactly the same way that we differentiate scalar functions. For a time-dependent vector →a(t), the derivative ˙→a(t) is:
Vector derivative definition.#rvc‑ed
˙→a(t)=ddt→a(t)=limΔt→0→a(t+Δt)−→a(t)Δt
Note that vector derivatives are a purely geometric concept. They don't rely on any basis or coordinates, but are just defined in terms of the physical actions of adding and scaling vectors.
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Increment: | Δt= 2 s |
Time: | t= 0 s |
Vector derivatives shown as functions of t and Δt. We can hold t fixed and vary Δt to see how the approximate derivative Δ→a/Δt approaches ˙→a. Alternatively, we can hold Δt fixed and vary t to see how the approximation changes depending on how →a is changing. #rvc‑fd
We will use either the dot notation ˙→a(t) or the full derivative notation d→a(t)dt, depending on which is clearer and more convenient. We will often not write the time dependency explicitly, so we might write just ˙→a or d→adt.
Derivatives and vector “positions” #rvc‑sp
When thinking about vector derivatives, it is important to remember that vectors don't have positions. Even if a vector is drawn moving about, this is irrelevant for the derivative. Only changes to length and direction are important.
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Movement: bounce stretch circle twist slider rotate vertical fly
Vector derivatives for moving vectors. Vector movement is irrelevant when computing vector derivatives. #rvc‑fp
Derivatives in components #rvc‑sc
In a fixed basis we differentiate a vector by differentiating each component:
Vector derivative in components#rvc‑ec
˙→a(t)=˙a1(t)ˆı+˙a2(t)ˆȷ+˙a3(t)ˆk
Derivation
Warning: Differentiating each component is only valid if the basis is fixed. #rvc‑wc
Time: | t= 0 s |
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Basis: | ˆı,ˆȷ ˆu,ˆv |
The vector derivative decomposed into components. This demonstrates graphically that each component of a vector in a particular basis is simply a scalar function, and the corresponding derivative component is the regular scalar derivative. #rvc‑fc
Differentiating vector expressions #rvc‑se
We can also differentiate complex vector expressions, using the sum and product rules. For vectors, the product rule applies to both the dot and cross products:
Product rule for dot-product derivatives.#rvc‑ep
ddt(→a⋅→b)=˙→a⋅→b+→a⋅˙→b
Product rule for cross-product derivatives.#rvc‑ex
ddt(→a×→b)=˙→a×→b+→a×˙→b
Example Problem: Differentiating vector product expressions. #rvc‑xe
The chain rule also applies to vector functions. This is helpful for parameterizing vectors in terms of arc-length s or other quantities different than time t.
Chain rule for vectors.#rvc‑er
ddt→a(s(t))=d→ads(s(t))dsdt(t)=d→ads˙s
Example Problem: Chain rule. #rvc‑er
Did you know?#rvc‑il
Gottfried Leibniz, one of the inventors of calculus, got the product rule wrong [Child, 1920, page 100; Cirillo, 2007]. In modern notation he computed the example ddx(x2+bx)(cx+d)=(2x+b)c and he stated that in general it was obvious that ddx(fg)=dfdxdgdx. He later realized his error and corrected it [Cupillari, 2004], but at least we know that product rules are tricky and not obvious, even for someone smart enough to invent calculus.
References
- J. M. Child. The Early Mathematical Manuscripts of Leibniz. Open Court Publishing, 1920. (Google ebook, local copy).
- M. Cirillo. Humanizing Calculus. The Mathematics Teacher, 101(1):23–27, 2007. (NCTM version, local copy)
- A. Cupillari. Another look at the rules of differentiation. Primus: Problems, Resources, and Issues in Mathematics Undergraduate Studies, 14(3):193–200, 2004. DOI: 10.1080/10511970408984087.
Changing lengths and directions #rvc‑sd
Two useful derivatives are the rates of change of a vector's length and direction:
Derivative of vector length.#rvc‑el
˙a=˙→a⋅ˆa
Derivation
Derivative of vector direction.#rvc‑eu
˙ˆa=1aComp(˙→a,→a)
Derivation
An immediate consequence of the derivative of direction formula is that the derivative of a unit vector is always orthogonal to the unit vector:
Derivative of unit vector is orthogonal.#rvc‑eu
˙ˆa⋅ˆa=0
Derivation
Recall that we can always write a vector as the product of its length and direction, so →a=aˆa. This gives the following decomposition of the derivative of →a.
Vector derivative decomposition.#rvc‑em
˙→a=˙aˆa⏟Proj(˙→a,→a)+a˙ˆa⏟Comp(˙→a,→a)
Derivation
Integrating vector functions #rvc‑si
The Riemann-sum definition of the vector integral is:
Vector integral.#rvc‑ei
∫t0→a(τ)dτ=limN→∞N∑i=1→a(τi)Δτ⏟→SNτi=i−1NΔτ=1N
In the above definition →SN is the sum with N intervals, written here using the left-hand edge τi in each interval.
Time: | t= 0 s |
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Segments: | N= 1 |
Integral of a vector function →a(t), together with the approximation using a Riemann sum. #rvc‑fi
Just like vector derivatives, vector integrals only use the geometric concepts of scaling and addition, and do not rely on using a basis. If we do write a vector function in terms of a fixed basis, then we can integrate each component:
Vector integral in components.#rvc‑et
∫t0→a(τ)dτ=(∫t0a1(τ)dτ)ˆı+(∫t0a2(τ)dτ)ˆȷ+(∫t0a3(τ)dτ)ˆk
Derivation
Warning: Integrating each component is only valid if the basis is fixed. #rvc‑wi
Example Problem: Integrating a vector function. #rvc‑xi
Warning: The dummy variable of integration must be different to the limit variable. #rvc‑wd